In the study of Chemistry, we encounter very large and very small numbers. For example, the size of a molecule or thin layer of molecules is very, very small and difficult to measure directly.
In order to “measure” such numbers, we often use one measured quantity and another known property to calculate the unknown value. In this experiment, we will use this approach to calculate the thickness of various types of aluminum foil, a quantity whose direct experimental determination is not possible with common laboratory equipment.
By measuring the mass of a piece of foil and by calculating its area, the thickness of the foil can be obtained from these quantities and the known density of aluminum, 2.70 g/cm3.
Show your work and express answers with the correct number of significant figures and proper units.
1. Find the volume of a block with the dimensions L = 9.16 cm, W = 3.65 cm, H = 1.01 cm.
2. If the density of a substance is 0.734 g/cm3 and the volume of the substance is 21.45 cm3, find its mass.
3. Determine the thickness of a rectangular piece of paper with an area of 45.3 cm2 and a volume of 7.7 X 10-3 cm3. (Hints: rewrite the formula for volume as length X width X thickness. Also, what is the formula for area?)
1. Cut 2 rectangular pieces of regular aluminum foil and 2 rectangular pieces of heavy-duty aluminum foil. Be sure that the dimensions are at least 10 cm on each side. Record the type (e.g. Reynolds Heavy Duty, Schnucks Regular) of each piece of aluminum foil in the data table below.
2. Using the ruler, carefully measure the length and width of each piece of foil. Record the measurements in your data table. Take care to record your measurements with the appropriate number of significant figures. Record the length and width of each piece of aluminum foil in the data table below.
3. Using a balance, find...